20.8 - Property VI

In the previous section, we saw the fifth property of determinants. In this section, we will see the sixth property.

Property VI
This can be written in 6 steps:
1. Let Δ = $\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3     }    \\
b_1    &{    b_2    }    &{    b_3    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right | $.
2. Pick any two rows, say R₂ and R₃.
Apply R₂ → R₂ + k R₃

• Then we get a new determinant Δ₁

Δ₁ = $\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3     }    \\
b_1 + k c_1    &{    b_2 + k c_2    }    &{    b_3 + k c_2    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right | $.

• Note that, when we apply R₂ → R₂ + k R₃, the row R₂ is affected. The other row R₃ remains the same.

3. We can apply property 5. That is., we can split Δ₁ and write it as the sum of two determinants:

$\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3     }    \\
b_1 + k c_1    &{    b_2 + k c_2    }    &{    b_3 + k c_2    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right | ~=~\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3    }    \\
b_1    &{    b_2    }    &{    b_3    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right |~+~\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3    }    \\
k c_1    &{    k c_2    }    &{    k c_3    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right |$

4. In the above equation, consider the second determinant in the R.H.S. All the elements of R₂ are proportional to the corresponding elements in R₃ in the same ratio k:1. So by applying property IV, this determinant is zero.

5. So the equation in (3) becomes:
$\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3     }    \\
b_1 + k c_1    &{    b_2 + k c_2    }    &{    b_3 + k c_2    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right | ~=~\left |\begin{array}{c}                           
a_1     &{    a_2     }    &{    a_3    }    \\
b_1    &{    b_2    }    &{    b_3    }    \\
c_1    &{    c_2    }    &{    c_3    }    \\
\end{array}\right |~+~0$

• That is: Δ₁ = Δ

6. We can write:
If we apply the operation R_i → R_i + k R_j, the value of the determinant remains same.
7. This is true for columns also. We can write:
If we apply the operation C_i → C_i + k C_j, the value of the determinant remains same.
8. When we apply R_i → R_i + k R_j, the row affected is R_i . The row R_j is not affected.
• In one step, we must not use R_i in another operation.
9. A rule similar to (8) can be written for columns also.

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Now we have a clear understanding about property VI. Let us see an example.

• Show that $\left|\begin{array}{r}                           
a    &{    a+b    }    &{    a+b+c    }    \\
2a    &{    3a+2b    }    &{    4a+3b+2c    }    \\
3a    &{    6a+3b    }    &{    10a+6b+3c    }    \\
\end{array}\right|~=~a^3$.
Solution:
 

◼ Remarks:
• 2, 3 and 4 (magenta color): Here we apply property VI
• 5(magenta color): Here we expand along C₁

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In the next section, we will see some solved examples.

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